Evaluating a Summation with a binomial Problem:

Evaluate for $n=11$$$\begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} $$

Sorry for this odd question. I saw this formula here on MSE which is quite helpful for a question I need to solve ($\sin^{4n}\frac{\pi}{4n} + \cos^{4n} \frac{\pi}{4n})$ for given values of $n$. Unfortunately I don't know how to evaluate this formula manually. I was thus hoping to use Wolfram Alpha to evaluate this for different values of $n$ but was unable to enter it correctly. I would be truly grateful if somebody would kindly show me how to input this formula into Wolfram Alpha or solve it for $n=11$. Many thanks in advance.
 A: With some trivial manipulation it straighforward to check that we just have to compute:
$$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\sum_{r=0}^{2n}\binom{4n}{2r}\exp\left(8r\cdot\frac{2\pi i}{4n}\right).\tag{1} $$
Since:
$$ \sum_{r=0}^{2n}\binom{4n}{2r} z^{2r} = \frac{1}{2}\sum_{k=0}^{4n}\binom{4n}{k}\left(z^k+(-z)^k\right)=\frac{(1+z)^{4n}+(1-z)^{4n}}{2}\tag{2}$$
by taking $\omega=\exp\frac{2\pi i}{n}$ we have:
$$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\left(\frac{(1+\omega)^{4n}+(1-\omega)^{4n}}{2}\right)\tag{3}$$
then, since $1\pm \omega = 2\cos\left(\frac{\pi}{n}\right)\exp\left(\pm\frac{\pi i}{n}\right)$,
$$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n} = \frac{1}{2}\left(2\cos\frac{\pi}{n}\right)^{4n}\left(\cos(4\pi)+\cos(-4\pi)\right)=\color{red}{\left(2\cos\frac{\pi}{n}\right)^{4n}}.\tag{4}$$
A: $\displaystyle e^{ix}=\cos x+i\sin x\implies e^{-ix}=\cos(-x)+i\sin(-x)=\cos x-i\sin x$
$\displaystyle e^{ix}+e^{-ix}=2\cos x,e^{ix}-e^{-ix}=2i\sin x$
$\displaystyle \left(2i\sin x\right)^{4n}+\left(2\cos x\right)^{4n}$
$\displaystyle =(e^{ix}+e^{-ix})^{4n}+(e^{ix}-e^{-ix})^{4n}$
$\displaystyle =2\sum_{r=0}^{2n}\binom{4n}{2r}(e^{ix})^{4n-2r}(e^{-ix})^{2r}$
$\displaystyle\implies16^n\left(\sin^{4n}x+\cos^{4n}x\right)$
$\displaystyle =2\sum_{r=0}^{2n}\binom{4n}{2r}e^{i4x(n-r)}$
$\displaystyle 2^{4n-1}\left(\sin^{4n}x+\cos^{4n}x\right)=\binom{4n}{2n}+\sum_{r=0}^{n-1}\left(\binom{4n}{2r}e^{i4x(n-r)}+\binom{4n}{4n-2r}e^{-i4x(n-r)}\right)$
As $\binom NR=\binom N{N-R},$
$\displaystyle\implies2^{4n-1}\left(\sin^{4n}x+\cos^{4n}x\right)$
$\displaystyle =\binom{4n}{2n}+\sum_{r=0}^n\binom{4n}{2r}\{e^{i4x(n-r)}+e^{-i4x(n-r)}\}$
$\displaystyle\implies2^{4n-1}\left[\sin^{4n}x+\cos^{4n}x\right]=\binom{4n}{2n}+\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{4(n-r)x\}$
Set $x=\dfrac\pi{4n}$
Can you take it home from here?
A: if you meant $\cos(1-4n/2r)*\pi$, then using Maple I got 

it seems for me that you cannot simplify the answer.
If you meant  $\cos[(1-4n/2r)*\pi]$, then here is the result

